I cannot access a variable within a function as a variable:
public function fetchWhere($params) {
$resultSet = $this->select(function(Select $select) {
$select->where($params);
});
..
I get the error:
Undefined variable: params
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3 Answers
You need the use construct then to make the variable available/visible inside the function:
public function fetchWhere($params) {
$resultSet = $this->select(function(Select $select) use($params) {
$select->where($params);
});
}
You can pass even more than just one variable with this. Just separate other variables with a comma , like ... use($param1, $param2, ...) {.
Comments
Closures may also inherit variables from the parent scope. Any such variables must be passed to the use language construct. It is because of variable scope. Try with -
$resultSet = $this->select(function(Select $select) use($params) {
$select->where($params);
});
Inheriting variables from the parent scope is not the same as using global variables. Global variables exist in the global scope, which is the same no matter what function is executing. The parent scope of a closure is the function in which the closure was declared (not necessarily the function it was called from).
Comments
use "use", that let you use a variable as a global in the scope :
public function fetchWhere($params) {
$resultSet = $this->select(function(Select $select) use($params) {
$select->where($params);
});
..
